some1 smart? (1 Viewer)

[crammy90]

1) in a electron transfer reaction, how do you know when a solution is either the electrolyte of the salt bridge...like in rusting the fe2+ and OH- move through the film of water on the surface of the fe structure...is this a salt bridge as it carries the ions, or is it an electrolyte as there are dissolved salts to allow for the charge to migrate...hmm..does a salt bridge have to be an electrolte and vice versa?

2) when we put iron in naoh and hcl, naoh doesnt form actual visible rust but produces fe2+, as shown when we add indication )K ferricyancide stuff)...is there more or less fe2+ in naoh than the hcl...like has the same amount been produced but some fe2+ reacted in hcl so theres more fe2+ present in naoh

3) for the 2007 paper it has "dilute Copper sulphate solution".
how do we know when we are picking equations if this is cu+ or cu2+?
and is the reason the OH- oxidises due to this wording of "dilute"

thanks

 

[Azreil]

1) Um. Okay. The saltbridge has to involve an ionic solution. Normally CH3COOK or a similar solution is used to ensure no precipitate forms. In the rusting of iron, I think that water (with dissolved ions) acts as both a saltbridge and an electrolyte in a weird way.
2) NaOH retards rusting. HCl encourages rusting. This is because:
H2O + e- --> 1/2H2(g) + OH-
This is actually an equillibrium reaction. You add H+ (ie in an acidic environment) and it consumes OH- thereby pushing the position of equilibrium right. You add OH- and you're adding an excess of products hence it shifts left; retarding rust. So more Fe2+ is produced in the HCl as the redox reactions have to occur simultaneously.
3) Unless it states it, it's Cu(II).
I don't know what you mean by the second part sorry.

 

[crammy90]

2) NaOH retards rusting. HCl encourages rusting. This is because:
H2O + e- --> 1/2H2(g) + OH-

so this "equilibrium" and this equation only applies to when we have an inert anode OR an electrolyte whose cation has a lower reduction potential ONLY in an electrolytic cell (as the reaction is ve- potential)...so having an acidic electrolyte in an electrolytic cell would accellerate rusting (according to this equation)...but this equilibrium doesnt apply to normal galvanic cells as its ve-...right?

and just say we didnt put enough electricity in for this reaction to occur as the cathode, what reduction equation would be used?

and for that equation, how do we know when to use that or the one with water AND oxygen...because wouldnt the 1 with oxygen go in preference to just water reducing...or does the "bubbles" in the question imply there is no oxygen dissoved and thus we use the 1 which produces the gas h

 

[samwell]

so this "equilibrium" and this equation only applies to when we have an inert anode OR an electrolyte whose cation has a lower reduction potential ONLY in an electrolytic cell (as the reaction is ve- potential)...so having an acidic electrolyte in an electrolytic cell would accellerate rusting (according to this equation)...but this equilibrium doesnt apply to normal galvanic cells as its ve-...right?

and just say we didnt put enough electricity in for this reaction to occur as the cathode, what reduction equation would be used?

and for that equation, how do we know when to use that or the one with water AND oxygen...because wouldnt the 1 with oxygen go in preference to just water reducing...or does the "bubbles" in the question imply there is no oxygen dissoved and thus we use the 1 which produces the gas h

Depends merely on context. An electrolyte is a liquid that enables the flow of electrons and in most cases acids are not as gud conductors as bases are. Secondly using an acid would accelerate rusting and thus using it in any iron galvanic cell would be destructive. Thats y in restoration they use NaOH because it is a gud conductor and it also does not accelerate rusting

The electrolysis needs a minimum amount of voltage in order to occur. If not enough voltage is applied then no reaction will take place.

In all rusting related half reactions the one containing:
O2 + H2O + 4e- ---> 4OH-
If the question doesnt have anything to do with rusting then water does not have to go on with the half eqn with O2 and its reduction or oxidation will be prefered.

Hope this helps

 

[Felix Jones]

quite sam!!!
hey how you been bro, i'll see you tomorrow. lol.
good luck.

 

[samwell]

quite sam!!!
hey how you been bro, i'll see you tomorrow. lol.
good luck.

lol tc bro am okay just finishing off my prac revision. lol gud luck 4 2moro